tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We use the identity $\cot 	heta - 	an 	heta = 2 \cot 2	heta$,which implies $\cot 	heta = 	an 	heta + 2 \cot 2	heta$.Rearranging,we get $2

Vedclass · Accepted Answer

$\cot A$

Vedclass · Answer

(B) We use the identity $\cot 	heta - 	an 	heta = 2 \cot 2	heta$,which implies $\cot 	heta = 	an 	heta + 2 \cot 2	heta$.Rearranging,we get $2 \cot 2	heta = \cot 	heta - 	an 	heta$.Now,consider the expression $E = 	an A + 2 	an 2A + 4 	an 4A + 8 \cot 8A$.Using the identity $	an 	heta = \cot 	heta - 2 \cot 2	heta$,we have:$	an A = \cot A - 2 \cot 2A$Substituting this into the expression:$E = (\cot A - 2 \cot 2A) + 2 	an 2A + 4 	an 4A + 8 \cot 8A$This approach is complex,so let us simplify step-by-step using $	an 	heta = \cot 	heta - 2 \cot 2	heta$ repeatedly:$8 \cot 8A = 4 \cot 4A - 4 	an 4A$$E = 	an A + 2 	an 2A + 4 	an 4A + (4 \cot 4A - 4 	an 4A) = 	an A + 2 	an 2A + 4 \cot 4A$Now,$4 \cot 4A = 2 \cot 2A - 2 	an 2A$$E = 	an A + 2 	an 2A + (2 \cot 2A - 2 	an 2A) = 	an A + 2 \cot 2A$Finally,$2 \cot 2A = \cot A - 	an A$$E = 	an A + (\cot A - 	an A) = \cot A$.

$\tan A + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A = $

Similar Questions

If $540^{\circ} < A < 630^{\circ}$ and $|\cos A| = \frac{5}{13}$,then $\tan \frac{A}{2} \tan A = $

If $\tan \frac{\theta}{2} = t$,then $\frac{1 - t^2}{1 + t^2}$ is equal to

If $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$,find the values of $\sin \frac{x}{2}$,$\cos \frac{x}{2}$,and $\tan \frac{x}{2}$.

If $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then $\log \sec x = $

If $2\cos^2 \theta - 2\sin^2 \theta = 1$,then $\theta = \dots ^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module